Let $B=B(r)$ denote a ball of radius $r$ in $\Omega \subset \mathbb R^d$ and $$ u_B := \frac1{|B|}\int_B u \, dx. $$ The standard Sobolev-Poincaré inequality states that if $u \in W^{1,p}(\Omega)$, then $$ \|u-u_B\|_{L^\kappa(B)} \le Cr \|\nabla u\|_{L^p(B)} $$ for $$ \kappa = \frac{dp}{d-p}, \quad 1 < p < d. $$

My question is whether something similar is true for merely curl-integrable functions. More precisely, suppose $u \in L_{\rm loc}^p$ is divergence-free and $\text{curl } u \in L_{\rm loc}^p, p>1$. As pointed out below, a standard Poincaré inequality with the gradient replaced by the curl, $$ \|u-u_B\|_{L^\kappa(B)} \le Cr \|\text{curl }u\|_{L^p(B)}, $$ cannot hold in general. But I suspect there should be a corresponding result with some extra assumptions or with some additional term on the right hand side. Any references in to that direction are appreciated.

I am mostly interested in the case $d=2$ and $1<p<2$.

Edit: If it is easier, I would also be interested just in the case, where $\kappa$ is replaced by $p \in (1,2)$ for $d=2$.

  • 1
    $\begingroup$ I'm not sure that it will contain the answer, but the following paper treats a very similar subject: Note on explicit proof of Poincare inequality for differential forms on manifolds by Leonid Shartser (arXiv:1010.3356). $\endgroup$ May 29, 2015 at 17:30
  • $\begingroup$ Looks interesting. But figuring out whether this gives what I want seems a non-trivial task with my geometry skills. I was hoping for a more definite answer for the question. $\endgroup$ Jun 2, 2015 at 7:48
  • $\begingroup$ Do you mean Friedrich's inequality? $\endgroup$
    – username
    Sep 23, 2015 at 14:30
  • $\begingroup$ To best of my knowledge Friedrich's inequality requires that the function is in some Sobolev space (i.e. all the partial derivatives, up to some order, are integrable). I was hoping to obtain something for merely curl-integrable functions (say, in 2D). $\endgroup$ Oct 2, 2015 at 8:12

1 Answer 1


Counterexample $(x,-y)$, or am I misunderstanding something here.

  • $\begingroup$ No, I think this is a valid counter-example to the curl-Poincaré inequality. However, then my question is that how should the result be modified to get something reasonable out of it. Somehow I feel there should be a corresponding result also in this setting. But maybe I am wrong. $\endgroup$ Jun 7, 2015 at 13:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.